Nodal Circuit Analysis Using KCL. Most useful for when we have mostly current sources Node analysis uses KCL to establish the currents

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1 Nodal Circuit Aalysis Usi KCL Most useful for whe we have mostly curret sources Node aalysis uses KCL to establish the currets Procedure () Choose oe ode as the commo (or datum) ode Number (label) the odes Desiate a voltae for each ode umber Each ode voltae is with respect to the commo or datum ode Number of odes used umber of odes - Note: umber of odes braches b- Thus less equatios with ode aalysis tha mesh aalysis () For each ode write the KCL for curret flows i each ode Use differeces i the ode voltaes to calculate currets Assume the curret directios ad write the KCL eerally assume the ode is a positive relative to all others Curret directios differet for same brach i each ode Ofte better to use coductace equatios () Solve the equatios for the ode voltaes et currets i each brach from the voltae differeces

2 Example Nodal Circuit Aalysis Cosider the ode, loop circuit below () Setti the base ode, ad ode voltaes Set the commo ode to roud Label voltaes o the others () For each ode write KCL for curret flows For ode, Defii the curret directios for as ito the ode Use differeces i the ode voltaes for currets Collect each voltae ito oe term ( ) Or i the coductace form ( ) ( )

3 Example Nodal Circuit Aalysis cotiued For ode, Defii the curret directios for as ito the ode This meas curret i differet from ode ( ) Or if chai to coductace form ( ) ( ) Now have two equatios ad ukows ( s) Could solve this alebraically stead use the umerical methods

4 Example Nodal Circuit Aalysis Putti the equatios i umerical form for ode : For ode : 000 ( 0.005) ( ) 0. 0 ( ) ( ) Usi substitutio method for ode Thus usi ode for the solutio ( ) 000 ( ) ( ) ( )

5 Example Nodal Circuit Aalysis Cotiued Solvi for ode ( ) 0.0 8( ) Curret i resistaces is the directio of ode ( ) ma 4 ma ma

6 Nodal Aalysis eeral Equatios eeral the odal equatios have the form: LL LL util L L where ij brach coductace betwee "i"th ode ad "j"th ode ii all brach coductace see by "i"th ode Thus i the example circuit ( ) ( ) Ad the terms are ( )

7 Dummy Nodes ad oltae Sources How do we solve a Node circuits cotaii voltaes sources? A oltae source fully defies the ode voltae This creates a "dummy ode" or superode Creates a ode costrait equatio that defies the voltae Example: the circuit below has curret ad voltae sources But is fully defied by voltae source Use this costrait equatio to remove oe ukow This reduces the umber of equatios to solved by ode Thus elimiate the ukow curret of the voltae source Thus ode ca be elimiated ad ode becomes ( ) Thus ode ca be solved directly

8 Mesh Aalysis usi KL (EC 4) Most useful whe we have mostly voltae sources Mesh aalysis uses KL to establish the currets Procedure () Defie a curret loop Set a directio for each simple closed path Number of loops eeded umber of braches b- Loop currets ca overlap: ofte may possible combiatios Must cover all braches with the loop set Each loop is called a Mesh () For each mesh write the KL equatio for the loops Whe loop currets overlap: Add currets if i same directio Subtract currets if i opposite directio oltae sources add if i the directio of loop curret oltae sources subtract if opposite to the loop curret () Solve the simultaeous equatios for the loop currets et currets i each brach from the loop currets oltaes calculated from the currets

9 Example Mesh Aalysis of Circuit Simple two source etwork, with braches () Establish two mesh currets (other loops iored) Number of loops b- - () Now write the KL equatios For loop : ( ) 0 Or more commoly putti o the riht For loop : ( ) 0 Aai etti o the riht ( ) ( ) These are the basic equatios of the etwork

10 Example Mesh Aalysis of Circuit Cot'd (EC 4.5) Solvi these two equatios ad ukows Typically use substitutio methods for simple equatios Use matrix methods for more complex circuits First solvi the loop equatios for ( ) Usi substitutio methods Now substituti for i the loop equatio ( ) ( ) Solvi for ad brii everythi to a commo deomiator ( )( ) ( ) ( ) ( ) [ ] ( ) ( ) [ ] Much more difficult solvi if do everythi alebraically

11 Example Mesh Aalysis of Circuit Cot'd (EC 4.5) Cosider the specific circuit the ( ) [ ] ( ) ( 000) 000( 000) 000( 000) 000 ma Solvi for the ma The the curret throuh is ma The solvi for the voltae across the resistors Use curret throuh each resistor Now curret throuh each source ma ma Note: has curret ito side: thus it is bei chared Havi all s & s completely solves the circuit

12 Mesh Aalysis of Circuit: Matrix Solutios For solvi this usi matrices use umerical equatios For loop : For loop : ( ) ( 000) ( ) ( ) 000 This makes maipulatio easier Note: some calculators have multiple equatio/ukow solvers Alteratively solve usi matrixes (see EC appedix A) esistors become a x matrix Curret a x colum matrix oltae a x colum matrix [ ][ ] [ ] The solve the equatios by iverti the matrix [ ] [ ] [ ]

13 Matrix Method ad Spread Sheets Easy to use matrix method i Excel or Matlab or Maple Use mivert ad mmult array fuctios i Excel Create the ad matrix i a spreadsheet First ivert the matrix: select output cells with same array size Eter miverse( The select the matrix cells e miverse(b4:c5) The press cotrolshifteter (very importat) Does ot properly eter array if you do ot do that This creates iverse of matrix at desired locatio The eed to multiply iverse times colum: use mmult( Select output colum cells the comma Select - cells ad cells (e mmult(b8:c9,d8:d6) ) The press cotrolshifteter Here is example from previous pae

14 Mesh Aalysis eeral Equatios eeral the mesh equatios have the form: util r r r r r r LL LL r r r r r LL r where r ij total resistace i the "i"th mesh see by curret "j" r ii total resistace i the "i"th mesh see by the i th curret loop E. i the example circuit for loop The the matrix terms are ( ) ( ) r r This is the eeral form of the equatios/ukows Also the eeral matrix form

15 Dummy Meshes ad Curret Sources How do we do mesh circuits cotaii a curret source? A Curret source fully defies the mesh curret This creates a "dummy mesh" or supermesh : Creates a mesh costrait equatio that defies the currets E. ad source the circuit below (same as i the dummy ode) The is fully defied by the source Use this costrait equatio to remove the ukow curret educes the umber of equatios to solve by mesh Thus elimiates the ukow voltae of curret source Thus loop ca be elimiated ad loop becomes ( ) ( ) Thus loop ca be solved directly

16 Dual Networks Two etworks are Duals whe the have similar equatios For the dual of a mesh etwork () Write the mesh equatios () eplace the currets with voltaes ad vise versa () eplace the resistaces with coductaces Example for the mesh circuit example below The the dual circuit is ( ) ( ) ( ) ( ) Note: curret directio of is i loop directio